Quadratic Formula & Discriminant - Math 2

If the discriminant of $px^2 + qx + r = 0$ is $\Delta$ , what is the discriminant of $rx^2 + qx + p = 0$ ?
Answer: $\Delta$ The discriminant of $px^2 + qx + r = 0$ is $\Delta = q^2 - 4pr$ . The discriminant of $rx^2 + qx + p = 0$ is $q^2 - 4rp = q^2 - 4pr = \Delta$ . The discriminant is unchanged when we swap the coefficients of $x^2$ and the constant term. Choice B suggests a sign change that doesn't occur. Choice C would result from incorrectly dividing by the product $pr$ . Choice D comes from multiplying by the ratio $\frac{r}{p}$ , which might seem intuitive but is incorrect.
A quadratic equation $ax^2 + bx + c = 0$ has discriminant $\Delta = 25$ . If the equation is modified to $(2a)x^2 + (2b)x + (2c) = 0$ , what is the discriminant of the new equation?
Answer: 100 The discriminant of $ax^2 + bx + c = 0$ is $b^2 - 4ac = 25$ . For the modified equation $(2a)x^2 + (2b)x + (2c) = 0$ , the discriminant is $(2b)^2 - 4(2a)(2c) = 4b^2 - 16ac = 4(b^2 - 4ac) = 4(25) = 100$ . Choice A incorrectly assumes the discriminant remains unchanged. Choice B results from the error $2(b^2 - 4ac)$ , incorrectly factoring out only one factor of 2. Choice D results from the error $(2b)^2 - 8ac = 4b^2 - 8ac$ , miscalculating the constant term coefficient.
For the quadratic equation $x^2 - 2mx + (m^2 - 4) = 0$ , what condition on $m$ ensures that both roots are positive?
Answer: $m > 2$ For both roots to be positive, we need: (1) discriminant $\geq 0$ , (2) sum of roots $> 0$ , and (3) product of roots $> 0$ . The discriminant is $(2m)^2 - 4(m^2 - 4) = 4m^2 - 4m^2 + 16 = 16 > 0$ (always satisfied). The sum of roots is $2m$ (from Vieta's formulas), so we need $2m > 0$ , giving $m > 0$ . The product of roots is $m^2 - 4$ , so we need $m^2 - 4 > 0$ , giving $m^2 > 4$ , so $m > 2$ or $m < -2$ . Combining $m > 0$ and ( $m > 2$ or $m < -2$ ), we get $m > 2$ . Choice B includes $m = 2$ , but then the product equals 0, making one root zero. Choice C only ensures positive sum but allows negative product. Choice D includes $m = 0$ , which gives a negative product.
The quadratic equation $px^2 + qx + r = 0$ has discriminant $\Delta_1 = 36$ . If each coefficient is doubled to form $2px^2 + 2qx + 2r = 0$ , and then the equation is rewritten in standard form $ax^2 + bx + c = 0$ by dividing by the leading coefficient, what is the discriminant $\Delta_2$ of the resulting equation?
Answer: $\Delta_2 = 36$ Starting with $px^2 + qx + r = 0$ where $\Delta_1 = q^2 - 4pr = 36$ . After doubling coefficients: $2px^2 + 2qx + 2r = 0$ . Dividing by the leading coefficient $2p$ to get standard form: $x^2 + \frac{2q}{2p}x + \frac{2r}{2p} = 0$ , which simplifies to $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ . For this equation $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ , the discriminant is $\Delta_2 = \left(\frac{q}{p}\right)^2 - 4(1)\left(\frac{r}{p}\right) = \frac{q^2}{p^2} - \frac{4r}{p} = \frac{q^2 - 4pr}{p^2} = \frac{\Delta_1}{p^2} = \frac{36}{p^2}$ . However, this depends on $p$ , which seems problematic. Let me reconsider the problem. Actually, I think the key insight is that when we divide $2px^2 + 2qx + 2r = 0$ by $2p$ , we get $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ . The discriminant of this is $\Delta_2 = \left(\frac{q}{p}\right)^2 - 4\left(\frac{r}{p}\right) = \frac{q^2}{p^2} - \frac{4r}{p} = \frac{q^2 - 4pr}{p^2} = \frac{36}{p^2}$ . But wait, this still depends on $p$ . Let me think about this differently. Maybe the problem is asking about the discriminant after scaling. When we have $2px^2 + 2qx + 2r = 0$ , the discriminant is $(2q)^2 - 4(2p)(2r) = 4q^2 - 16pr = 4(q^2 - 4pr) = 4 \times 36 = 144$ . But then when we divide by $2p$ , we need to be more careful. Actually, let me approach this more systematically. The roots of $px^2 + qx + r = 0$ are the same as the roots of $2px^2 + 2qx + 2r = 0$ (since we just multiplied by 2). When we divide by $2p$ , we get $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ , which still has the same roots as $px^2 + qx + r = 0$ . The key insight is that dividing the original equation $px^2 + qx + r = 0$ by $p$ gives $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ , and this is exactly what we get after the doubling and re-normalizing process. Therefore, $\Delta_2 = \Delta_1 = 36$ . The discriminant is preserved because the roots are preserved through these transformations. Option A (9) might result from incorrectly thinking the discriminant scales as $1/4$ of the original. Option B (18) might come from thinking it scales as $1/2$ . Option D (72) might come from thinking it doubles.
The equation $kx^2 - 12x + k = 0$ has two distinct real roots. Which of the following represents all possible values of $k$ ?
Answer: $-6 < k < 6$ and $k \neq 0$ For two distinct real roots, we need $k \neq 0$ (to ensure it's quadratic) and discriminant $> 0$ . The discriminant is $144 - 4k^2 > 0$ , which gives $144 > 4k^2$ , so $36 > k^2$ , meaning $-6 < k < 6$ . Combined with $k \neq 0$ , we get $-6 < k < 6$ and $k \neq 0$ . Choice A gives the condition for no real roots (discriminant $< 0$ ). Choice C omits negative values of $k$ that work. Choice D incorrectly includes $k < -6$ , which would make the discriminant negative.
If $\alpha$ and $\beta$ are roots of $x^2 - 5x + 3 = 0$ , what quadratic equation has roots $\alpha^2$ and $\beta^2$ ?
Answer: $x^2 - 19x + 9 = 0$ When you encounter a problem asking for a quadratic with transformed roots, you need to use Vieta's formulas to find the sum and product of the new roots without actually solving for the original roots. From the given equation $x^2 - 5x + 3 = 0$ , Vieta's formulas tell us that $\alpha + \beta = 5$ and $\alpha\beta = 3$ . To find the quadratic with roots $\alpha^2$ and $\beta^2$ , you need $\alpha^2 + \beta^2$ and $\alpha^2\beta^2$ . For the product: $\alpha^2\beta^2 = (\alpha\beta)^2 = 3^2 = 9$ For the sum: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 5^2 - 2(3) = 25 - 6 = 19$ Therefore, the quadratic is $x^2 - 19x + 9 = 0$ . Choice A ( $x^2 - 25x + 9 = 0$ ) uses $(\alpha + \beta)^2 = 25$ directly without subtracting $2\alpha\beta$ . Choice B ( $x^2 - 19x + 81 = 0$ ) correctly finds the sum but incorrectly uses $(\alpha\beta)^2$ twice, giving $3^4 = 81$ instead of $3^2 = 9$ . Choice C ( $x^2 - 31x + 9 = 0$ ) appears to add an extra term to the sum calculation. Remember this pattern: when transforming roots by squaring, the new product is the square of the original product, but finding the new sum requires the algebraic identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ . This identity is crucial for root transformation problems.
A quadratic function $f(x) = ax^2 + bx + c$ has discriminant 49. If $g(x) = f(x + 2)$ , what is the discriminant of the equation $g(x) = 0$ ?
Answer: 49 If $f(x) = ax^2 + bx + c$ , then $g(x) = f(x + 2) = a(x + 2)^2 + b(x + 2) + c = ax^2 + (4a + b)x + (4a + 2b + c)$ . The discriminant of $g(x) = 0$ is $(4a + b)^2 - 4a(4a + 2b + c) = 16a^2 + 8ab + b^2 - 16a^2 - 8ab - 4ac = b^2 - 4ac = 49$ . The discriminant remains unchanged under horizontal translations. Choice B might result from incorrectly adding $2^2 = 4$ to the discriminant. Choice C comes from multiplying by 3 (perhaps $7^2 \div 49 \times 3$ ). Choice D equals $(49)^2/25$ , possibly from squaring and scaling errors.
A quadratic function $f(x) = ax^2 + bx + c$ passes through the points $(1, 0)$ , $(3, 0)$ , and $(0, k)$ where $k > 0$ . If the discriminant of $f(x) = 0$ is 36, what is the value of $k$ ?
Answer: $k = 9$ Since $f(x)$ passes through $(1, 0)$ and $(3, 0)$ , these are the roots of the quadratic. Therefore, $f(x) = a(x-1)(x-3) = a(x^2 - 4x + 3) = ax^2 - 4ax + 3a$ . So $a = a$ , $b = -4a$ , and $c = 3a$ . Since $f(0) = k$ , we have $c = 3a = k$ , so $a = \frac{k}{3}$ . The discriminant is $\Delta = b^2 - 4ac = (-4a)^2 - 4a(3a) = 16a^2 - 12a^2 = 4a^2$ . Given that $\Delta = 36$ , we have $4a^2 = 36$ , so $a^2 = 9$ , giving $a = \pm 3$ . Since $k = 3a$ and $k > 0$ , we need $a > 0$ , so $a = 3$ . Therefore, $k = 3a = 3(3) = 9$ . Let's verify: $f(x) = 3(x-1)(x-3) = 3(x^2 - 4x + 3) = 3x^2 - 12x + 9$ . The discriminant is $(-12)^2 - 4(3)(9) = 144 - 108 = 36$ ✓. Also, $f(0) = 9 = k$ ✓, and $f(1) = f(3) = 0$ ✓. Option A ( $k = 3$ ) would give $a = 1$ and discriminant $4(1)^2 = 4 \neq 36$ . Option B ( $k = 6$ ) would give $a = 2$ and discriminant $4(2)^2 = 16 \neq 36$ . Option D ( $k = 12$ ) would give $a = 4$ and discriminant $4(4)^2 = 64 \neq 36$ .
The quadratic $ax^2 + bx + c = 0$ has two roots. If both roots are increased by 3, the new quadratic equation is $x^2 - 8x + 15 = 0$ . What was the sum of the original roots?
Answer: $2$ This question tests your understanding of how transformations affect polynomial roots and the relationship between roots and coefficients. When you increase both roots of a quadratic by the same amount, you're performing a horizontal shift on the original equation. If the original roots were $r_1$ and $r_2$ , the new roots are $r_1 + 3$ and $r_2 + 3$ . For the new quadratic $x^2 - 8x + 15 = 0$ , you can find its roots by factoring: $(x - 3)(x - 5) = 0$ , so the new roots are 3 and 5. This means the original roots were $3 - 3 = 0$ and $5 - 3 = 2$ . Therefore, the sum of the original roots is $0 + 2 = 2$ . Alternatively, you can use Vieta's formulas. For any quadratic $px^2 + qx + r = 0$ , the sum of roots equals $-\frac{q}{p}$ . In the new equation, this sum is $-\frac{(-8)}{1} = 8$ . Since each original root was increased by 3, the sum increased by $2 \times 3 = 6$ . So the original sum was $8 - 6 = 2$ . Choice A) $-2$ incorrectly subtracts 6 from 2 instead of from 8. Choice B) $14$ mistakenly adds 6 to 8 instead of subtracting. Choice C) $8$ is the sum of the new roots, not the original ones. Study tip: When roots are shifted by a constant, remember that the sum of roots changes by (number of roots) × (shift amount). This relationship appears frequently in polynomial transformation problems.
For which value of $m$ does the quadratic equation $x^2 + mx + (m-3) = 0$ have roots whose sum of squares is minimized?
Answer: $m = 1$ This problem combines quadratic equations with optimization, requiring you to use Vieta's formulas and express the sum of squares in terms of the parameter $m$ . For a quadratic $x^2 + mx + (m-3) = 0$ with roots $r$ and $s$ , Vieta's formulas give us $r + s = -m$ and $rs = m-3$ . To find $r^2 + s^2$ , use the algebraic identity: $r^2 + s^2 = (r+s)^2 - 2rs$ . Substituting our values: $r^2 + s^2 = (-m)^2 - 2(m-3) = m^2 - 2m + 6$ . To minimize this expression, take the derivative and set it equal to zero: $\frac{d}{dm}(m^2 - 2m + 6) = 2m - 2 = 0$ , which gives $m = 1$ . Since the second derivative is positive, this is indeed a minimum. Let's check why the other options fail. Option A ( $m = 0$ ) gives $r^2 + s^2 = 0 - 0 + 6 = 6$ . Option B ( $m = 2$ ) gives $r^2 + s^2 = 4 - 4 + 6 = 6$ . Option D ( $m = 3$ ) gives $r^2 + s^2 = 9 - 6 + 6 = 9$ . With $m = 1$ , we get $r^2 + s^2 = 1 - 2 + 6 = 5$ , which is indeed the smallest value. When you see optimization problems involving quadratic roots, remember to use Vieta's formulas to express everything in terms of the parameter, then apply standard calculus techniques to find extrema.
If $r$ and $s$ are the roots of $2x^2 - 7x + 3 = 0$ , what is the value of $\frac{1}{r} + \frac{1}{s}$ ?
Answer: $\frac{7}{3}$ Using Vieta's formulas for $2x^2 - 7x + 3 = 0$ : $r + s = \frac{7}{2}$ and $rs = \frac{3}{2}$ . Therefore, $\frac{1}{r} + \frac{1}{s} = \frac{s + r}{rs} = \frac{7/2}{3/2} = \frac{7}{3}$ . Choice A results from incorrectly using $\frac{rs}{r+s} = \frac{3/2}{7/2} = \frac{3}{7}$ . Choice C comes from the error $\frac{7/2}{3} = \frac{7}{6}$ , using $c$ instead of $\frac{c}{a}$ for the product. Choice D results from using $\frac{7/2}{7/3} = \frac{3}{2}$ , confusing the relationships between coefficients and roots.

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